$A$ rod $AB$ of length $15 \ cm$ rests between two coordinate axes such that the end point $A$ lies on the $x-$axis and the end point $B$ lies on the $y-$axis. $A$ point $P(x, y)$ is taken on the rod such that $AP = 6 \ cm$. Show that the locus of $P$ is an ellipse.

(N/A) Let $AB$ be the rod making an angle $\theta$ with the $x-$axis as shown in the figure. Let $P(x, y)$ be a point on the rod such that $AP = 6 \ cm$.
Since $AB = 15 \ cm$,we have $PB = AB - AP = 15 - 6 = 9 \ cm$.
From $P$,draw $PQ$ and $PR$ perpendiculars on the $y-$axis and $x-$axis,respectively.
In $\Delta PBQ$,$\cos \theta = \frac{PQ}{PB} = \frac{x}{9}$,so $x = 9 \cos \theta$.
In $\Delta PRA$,$\sin \theta = \frac{PR}{AP} = \frac{y}{6}$,so $y = 6 \sin \theta$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we substitute $\cos \theta = \frac{x}{9}$ and $\sin \theta = \frac{y}{6}$:
$(\frac{x}{9})^2 + (\frac{y}{6})^2 = 1$
$\frac{x^2}{81} + \frac{y^2}{36} = 1$.
This is the equation of an ellipse. Thus,the locus of $P$ is an ellipse.